In the given circuit, a 20 V ideal battery is connected across the...

An ideal ammeter has zero resistance, so it acts as a short circuit between nodes Q and R. This means nodes Q and R are at the same potential.

The circuit can be simplified as two parallel combinations in series. The $2\ \Omega$ and $4\ \Omega$ resistors are in parallel, and the $6\ \Omega$ and $3\ \Omega$ resistors are in parallel.

The equivalent resistance of the first parallel combination (between P and Q/R) is:
$R_1 = \dfrac{2 \times 4}{2 + 4} = \dfrac{8}{6} = \dfrac{4}{3}\ \Omega$

The equivalent resistance of the second parallel combination (between Q/R and S) is:
$R_2 = \dfrac{6 \times 3}{6 + 3} = \dfrac{18}{9} = 2\ \Omega$

The total equivalent resistance of the circuit is:
$R_{eq} = R_1 + R_2 = \dfrac{4}{3} + 2 = \dfrac{10}{3}\ \Omega$

The total current drawn from the battery is:
$I = \dfrac{V}{R_{eq}} = \dfrac{20}{10/3} = 6\text{ A}$

The voltage drop across the first parallel combination is:
$V_1 = I \times R_1 = 6 \times \dfrac{4}{3} = 8\text{ V}$

The current flowing through the $2\ \Omega$ resistor (from P to Q) is:
$I_{PQ} = \dfrac{V_1}{2} = \dfrac{8}{2} = 4\text{ A}$

The voltage drop across the second parallel combination is:
$V_2 = I \times R_2 = 6 \times 2 = 12\text{ V}$

The current flowing through the $6\ \Omega$ resistor (from Q to S) is:
$I_{QS} = \dfrac{V_2}{6} = \dfrac{12}{6} = 2\text{ A}$

Applying Kirchhoff's Current Law at node Q, the current entering the node must equal the current leaving it. Let $I_A$ be the current flowing through the ammeter from Q to R:
$I_{PQ} = I_{QS} + I_A$
$4 = 2 + I_A$
$I_A = 2\text{ A}$

Thus, the reading of the ammeter is $2\text{ A}$.

Answer: $2\text{ A}$